What is the extreme f(x)=(2x^2+18)/x ?

The extreme f(x)=(2x^2+18)/x is Maximum(-3,-12),Minimum(3,12)

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extreme f(x)=(2x^2+18)/x

extreme points

f (x) = \frac{2 x ^{2} + 18}{x}

Maximum (- 3, - 12), Minimum (3, 12)

Solution steps

f^{'} (x) = \frac{2 x ^{2} - 18}{x ^{2}}

Find intervals:

Increasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

\frac{2 x ^{2} + 18}{x} : - 12

Maximum (- 3, - 12)

Plug

x = 3

into

\frac{2 x ^{2} + 18}{x} : 12

Minimum (3, 12)

Maximum (- 3, - 12), Minimum (3, 12)

f (x) = - x^{2} - 15.9 x - 57

f (x, y) = x^{2} + y^{2} - 4 x + 2 y

f (x) = \frac{- x ^{2} - 10 x - 13}{x + 9}

x^{3} - 9 x^{2} + 1

f (x) = \frac{- 24 x ^{2} - 20 x + 24}{( x ^{2} + 1 ) ^{2}}

What is the extreme f(x)=(2x^2+18)/x ?
The extreme f(x)=(2x^2+18)/x is Maximum(-3,-12),Minimum(3,12)